TJNGKATAN CAW ANG AN PU LAU PIN ANG - TUTOR MANSOR · nama tjngkatan maj lis pengetua sekolah...

NAMA

T JNGKATAN

MAJ LI S PENGETUA SEKO LAH 1VIALA YSIA CAW ANG AN PU L AU PIN ANG

MODUL LATIHAN BERFOKUS SPM 2015 ADDITIONAL MA THEMATICS Kertas J Ogos 2 jam

3472/1

D ua j am

J ANGAN BUKA KE RTAS SOALAN INI SEHINGGA DIBERITAHU

1. T11/iska11 nama da11 tingka tan rmda pada m a11ga11 yang disediaka11.

2. Ker/as soalan i11i adalah da/am dwibalwsa.

3: .Soala11 da/0111 bahasa /11ggeris 111e11da/111/11i ".w ala11 yang sepada11 dalm11 balwsa Me/ayu.

4. Ca/011 dibe11arka11 111e11j mvab kese/11mhr111 atau sebahagia11 soa/a11 sama ada da/am bahasa !11ggeris a/au bahasa J\t/eloyu.

5. Ca/011 dikehe11daki 111e111baca maklumat di /Jala11w11 belaka11g kertas soa/m1 i11i.

U11t11k Kef{u1wm1 Pemeriksa

Soalan Markah Markah Pen uh Diperolehi

1 2 2 2 3 2 4 3 5 4 6 3 7 4 8 4 9 3 10 3 1L 4 12 2 13 3 14 3 15 4 16 3 17 3 18 4 19 4 20 4 21 3 22 4 23 3 24 2 25 4

J1JMLAH 80

Kerlas soalan ini mengandungi 24 halaman bercetak dan l halamau tidak bcrcetak.

3472/ J [Lihat ha lama n scbelah

tutormansor.wordpress.com

0.0

0.1

0.2

0.3

o..i ()5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6

l.'1

1.8

1.9

2.0

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

.l.O

2 3472/1

TlIE lJPPl<:R TAIL PROUAillLITY Q(z) FOR THE NORMAL DISTRIBUTION N(O, 1) I<EBARANGKALIAN IIUJUNG ATAS Q(z) BAGI TA BURAN NORMAL N(O, 1)

()

0.5000

0.4602

0.'1207

0.3821

0.3'116

<U085

0.27•1l

0.2420

0.2 119

0.1841

0.1 581

0.1357

0.1151

0.0968

0.0808

0.0668

0.0548

0.0•116

0.0359

0.0287

0.0228

IJ.01 79

0.0139

0.0107

0.00820

0.00621

0.00166

0.00.\47

0.00256

0.00187

0.00 135

I

.4%0 0

0

()

..1562

.4161{

.rnn

.3•109

0

0

0

(}

()

0

0

0

0

0

0

0

()

.3050

.2709

.2389

.2090

.1814

.1562

.1335

.11 31

.0951

.0793

.0655

.0537

.0136

.0351

.0281

.0222

.017•1

.0136

.0101

0

0

0

0

0

0

0

0

0 .00798

0 .00<>0 I

0 .0015.l

0 .00336

0 .00248

0 .00181

0 .00131

2

0.4920

0.4522

0.<11 29

0.3745

0.3372

0.101 5

0.2676

0.2358

0.2061

0.1788

0.1539

0.1314

0.111 2

0.0934

0.0778

0.0613

0.0526

0.0427

0.03•1 I

0.0274

0.0217

0.0170

0.0132

0.0102

0.00776

0.00587

0.001·10

0.00326

0.00240

0.00175

0.00126

0

Q(z) = J f( z) dz A

3472/l

3

0.4880

0.448.l

0.'1090

0.3707

0.3336

0.2981

0.2643

0.2327

0.2033

0.1762

0.1515

0.1292

0.1093

0.0918

0.0764

0.0630

0.0516

0,0418

0.0336

0.0268

0.0212

0.0166

0.0129

0.00990

0.00755

0.00570

0.00'127

O.OOJ 17

0.00233

0.00 169

0.00122

4 5 6 7

0.4840 0.4801 0.4761 0.472 1

0.4443 0.4404 0.<1364 0.4325

0.4052 0.<1013 0.3974 0.3936

0.3669 0.3632 0.3594 0.3557

0.3300 0.3264 0.3228 0.3192

0.29116 0.291 2 0.2877 0.2843

0.26 11 0.2578 0.2546 0.25 14

0.2296 0.2266 0.2236 0.2206

0.2005 0.1977 0.1949 0.1922

0. 1736 0.171 1 0.1685 0.1660

0. 1492 0. 1469 0. 1446 0.1423

0.1271 0.1251 0.1230 0.1210

0.1075 0.1056 0.1038 0.1020

0.0901 0.0885 0.0869 0.0853

0.071\9 0.0735 0.0721 0.0708

0.0618 0.0606 0.0594 0.0582

0.0505 0.0495 0.0485 0 .. 0475

0.0409 0.0401 0.0392 0.0384

0.0329 0.0322 0.0314 0.0307

0.0262 0.0256 0.0250 0.0244

0.0207 0.0202 0.0197 0.0192

0.0 162 0.0158 0.0154 0.0150

0.01 25 0.0122 0.0119 0.0116

0.00964 0.00939 0.009 14

0.00889

0.00734

0.00714 0.00695 0.00676

0.00554 0.00539 0.00523 0.00508

0.00415 0.00402 0.00391 0.00379

0.00307 0.00298 0.00289 0.00280

0.00226 0.002 19 0.00212 0.00205

0.00164 0.00159 0.00154 0.00149

0.001 18 0.00114 0.001 11 0.00107

/(z)

0 k

I 2 3 ti 5 6 8 9

1'.'1il1us I To/11k

0.4681 0.4641 4 8 12 16 20 24

0.4286 0.'1247 4 8 12 16 20 24

0.3897 O .. l859 4 8 12 15 19 23

0.3520 0.3483 4 7 II 15 19 22

0.3156 0.312 1 4 7 11 15 18 22

0.2810 0.2776 3 7 IO 14 17 20

0.2•183 0.245 1 3 7 IO 13 16 19

0.2177 0.2148 3 6 9 12 15 18

0.1894 0.1867 3 5 8 II l•I 16

0.1635 0.161 1 3 5 8 10 13 15

0.1401 0.1379 2 5 7 9 12 14

0.1190 0.1170 2 4 6 8 10 12

0.1003 0.0985 2 4 6 7 9 II

0.0838 0.0823 2 3 5 6 8 10

0.0694 0.068 1 I 3 4 6 7 8

0.0571 0.0559 I 2 4 5 6 7

0.0465 0.0455 I 2 3 4 5 6

0.0375 0.0367 I 2 3 4 •I 5

0.0301 0.0294 I I 2 3 4 4

0.0239 0.0233 I I 2 2 3 4

0.0188 0.0 183 0 I I 2 2 3

0.0146 0.0 143 0 I I 2 2 2

0.0113 0.0 11 0 0 I I I 2 2

0 I I I I 2

3 5 8 10 13 15

0.00866 0.00842 2 5 7 9 12 14

2 4 6 8 II 13

0.00657 0.00639 2 4 6 7 9 II

0.00494 0.00480 2 3 5 6 8 9

0.00368 0.00357 I 2 3 5 6 7

0.00272 0.00264 I 2 3 4 5 6

0.00 199 0.00193 I I 2 3 11 4

0.00 1<14 0.00139 0 I I 2 2 3

0.00 104 0.00100 0 I I 2 2 2

Example I Co11toh:

If X - N(O, I), then

JikaX - N(O, I), 111aka

P(X>2. l) = Q(2. I) = 0.0 179

7 8 s

28 32 31

28 32 31

27 31 3~

26 30 3<

25 29 3;

24 27 31

23 26 2S

21 21\ 27

19 22 25

18 20 23

16 19 21

II\ 16 18

13 15 17

II 13 14

10 II 13

8 10 II

7 8 9

6 7 8

5 6 6

4 5 5

3 4 4

3 3 4

2 3 3

2 2 2

18 20 23

16 16 21

15 17 19

13 15 17

II 12 14

9 9 10 I 7 8 9

5 6 6

3 4 4

3 3 4


3 3472/1

The fo llowing formulae may be helpful in answering the questions. The symbols given are the ones commonly used.

R1111111s-n11n11s berikut boleh 111e111bc111tu c111da 111e1?imvab soa/m1. Si111bol-si111bol yang diberi adalah yang biasa digunakan.

ALGEBRA

X = - b ± Jb 2 - 4ac 8

2a

2 a111 xa11 = a111+

11

9

3 a"' +a 11 = C/111

-11

10

4 (a"' )'= a11111

11 5 loga mn = loga m + log

11 n

12 6 I

111- 1 - I oga - OK, m oga n

n

7 log" 11111 = n loga m 13

CALCULUS KALKULUS

dy dv du y = uv, -= 11 - +v-

clr dx dx

4

du dv v--u-

u dy dx dx 2 y = - , = ~=----=-'"-\/ dx v2

3 dy dy du -=-X-dx du dx

5

3472/1

log b = loge b a loge a

7;, = a+ (n - l')c/

n S

11 = - [2a + (n - l)d]

2

T = ar11-

1

II

Su= a(r11 - 1)

= a(l - r 11

)

,. - I , _,.

a Ir! < I SW=--, 1- 1'

Area under a curve Luas di bawah /e11gk1111g

b

= J ydx or (atau)

"

b

= Jxdy (I

Volume of revolution Isi padu ki.rnm11

h

= J rr/ dr or (((/011) ti

" = f JCX2

l (}I

ti

, /' ;f; l

fLihat halam:111 schclah


.\- =

2 - L .fx y =--. I: .r

3

4

STATISTICS STATISTJJ(

7

8

9

3472/1

- L, w. J. I = I I

L, 111; n!

"P =- -,. (11 - r)!

n! "C =---

,. (11 - r)!r! " ~ ~L,(x,;'')' ~ ~L;;' _ -~'

10 P(Au B) =P(A)+P(B)- P(AnB)

4 11 P(X = r)="C,.p''q''_,. , p + q = I

5

6

2

[ 1 J - N - F

Ill = L + 2 . c J,,,

l = ~x lOO Qo

Distance I Jared<

=Jex, - x2)2 + CYi - Y2 )2

Midpoint I Titik tengali

. 1

_ (x, + X 2 Yi + Y2 J (.\,)) - 2 ' 2

12

13

14

GEOMETRY GEOMETRI

5

6

3 A point di viding a segment of a line Titik yang 111e111bahagi :matu te111bere11g garis

r 1

= (nx1 + 111x2 11y 1 + 111y2 J (. J ) ) Ill + /1 ' Ill + 11

4 Area of triangle I Luas segi tiga

= ~l(X1J'2 + X2Jl3 + X3.Y1 1 ) - (X2.Y1 + X3.Y2 + X1J13)1

3472/1

Mean / Mi11,11 = lljJ

a = J npq

X - p Z =--

CY

I t:. I = Jx2 + i

x! + y j ,. = - J x 2 + y 2


2

3

4

5

6

7

Arc length , s = r B Pm1jr111g lengkok, s = j B

l 2 Area of sector , A = - r B

2

i ·2e L11as sektor L = - J -, 2

sin 2A + cos2A = 1

sin 2 A+kos2A = 1

sec2A = I + tan2A sek 2 A = I + tan2 A

cosec2 A = 1 + cot2 A

kosek2 A = l + kot2 A

sin 2A = 2 sin A cos A sin 2A = 2 sin A kos A

cos2A = cos 2 A-sin2 A

= 2cos2 A - 1

= 1- 2sin 2 A

kos 2A = kos2 A- sin 2 A

= 2kos2 A-1

= l -2s in 2 A

3472/1

5 3472/1

TRIGONOMETRY TJUGONOMETRI

8

9

JO

11

12

sin( A± B) = sin A cos B ±cos A s in B sin (A ± B) =sin A kos H ± kos A sin B

cos(A ±B) = cos A cosB +sin A sin B

kos(A±B)= kosAkosB+si11 A sin B

_ (A B) tan A± tanB tan ± = -----1 + tan A tan B

2A 2 tanA

tan =-- -l - tan2 A

a b c = =

sin A sin B sin C

13 a2 = b 2 +c2 - 2bc cosA

a2 = b2 +c2 - 2bc kosA

14 Area of triangle I L11as segitiga

=_!_ab sin C 2

I Li hat lrnlaman scbclah tutormansor.wordpress.com

For Examiner 's

Use

1

[[ij

2

[[ij 0

3472/1

1

2

6

Answer all questions. Ja1vab semua soafa11.

Diagram l shows the relation between set P and set Q.

Rajah J 111e111111j11kka11 lwb1111ga11 a11tam set P da11 set Q.

State Nyataka11

z

Set P Diagram 1 Rajah I

Set Q

(a) the type of relat ion between set P and set Q, Jeni.\· //llb1111ga11 a11tam set P da11 set Q,

(b) the range of the relation. ju/a/ bagi lu1b1111ga11 itu.

Answer I Jawapa11: (a)

(b)

3472/l

[2 marks] [2 111arkah]

The function g is defined by g : x ---+ ~, x ~ 111. Find the va lue of 111 if g - 1 (6) = 4. x - 111

[2 marks]

F . / ' f< ;n b . 2 111 C . '{ . .. , - I (6) 4 , 1111gs1gCi 1ta npw11 se aga1 g : x ---+ - - , x ~ 111. an 111 m 111111w g = . x - 111

[2 111arkcil1] Answer I Jawapa11:


7 J 1172/l

3 In Diagram 3, the function g maps x onto y and the function/maps y onto z. Do/am Rl!fali 3,j/ 111gsi g 111e111eta/wn x kepada y da11.f11ngsi f111e111etaka11ykepada 1.

Determine Te11t11ka11

(a) .r-I (5),

.r --'8=--- Y·-"""'J __ z

1

(b) g - 1/ - 1(5).

Answer I Jmvapc111:

(a)

(b)

Diagram 3 Rlljali 3

i. 2 marks l [/. II I</ r/w /i J

4 The following information refers to the equation of the straight li ne y = I which is the tangent to the curve y = - x 2 + 6x - 3.

Jvfak/11111at berik111111en!f 11k kepacla persa111aa11 garis lums y = I iait11 ta11ge11 kepada /e11gkw1g y = - x 2 + 6x - 3.

Find the value of t.

Cari nilai I.

Answer I Jawapa11:

y = t y =- x 2 + 6x - 3

[3 111arks] l3 l//(/r/(((/J]

For rra111i11er·~

Uw

4

·~

( _) 3472/1 !Lihat ha la111a11 sl'l>clah


, .. (/,. 5 F:xaminer's

Use

5

w 0

3472/1

8 3472/1

Diagram 5 shows the curves y = 2(.r - I )2 - 11 and y = (x - m)2

- (11 - 4), where /11 and 11 are constants. Both the curves intersect the x- ax is at x = - I. Rajah 5 111e111111j11kka11 /e11gk1111g y = 2(.Y - I )2 - 11 da11 y = (x - 111)2 - (11 - 4), de11ga11 keadaa11111 da11 11 ada/ah pemalar. Ked11a-d11a le11gk1111g it11111e11yila11g paksi-x pada x =- 1.

Find Cari

y

Diagram 5 Rajah 5

(a) the value of m and of 11,

11i/ai 111 da1111ilai 11,

y = (x - 111)2

- (11 - 4)

2 y = 2(.r - l ) - 11

(b) the coordinates of the minimum point of each curve. koordi11at titik 111i11i11111111 bagi setiap le11gk1111g i/11.

Answer I Jawapa11:

(a)

(b)

[4marks] [ 4 111arka/J]


9 3472/1

6 Solve the equat ion .16 (2 x + 2

) = 8 x [3 marks] For Examiner's

. ,. + 2 ,. Se/esa1ka11 persa111aa11 16 (2 · ) = 8 · [3 markah] Use

Answer I Jawapa11:

7 Given that log3 x = lz and log3 y = k, express log9 [L] in terms of Ii and k. 27x

Diberi log3 x = Ii clan log3 y = k, 1111gkapka11 log9

Answer I Jawapa11:

3472/1

[ Y

2

] da/am seb11ta11 Ii da11 k. 27x

[4 marks] [ 4 mark ah]

7

cm 0

ILihat halaman sebelah


Fvr J::.rnmiuer '.~

Use

8

[Gj

0 3472/1

8

10 3472/1

The price of a medium cost house in Butterworth on I st April 2014 is RM 120,000. At that time Hassan lrns a saving of RM 100,000. The price of the house is expected to increase by 6 % every year while Hassan's saving is expected to increase by 8% every year. On the l st April of which year Hassan manage to purchase the house in cash?

Harga sebuah n111w/J kos seder/1011a di Butterworth pada I April 2014 adalah RM 120,000. Pada masa it11 Hassan 111e111p1111yai si111pa11a11 seba11yak RM 100,000. Harga m111a/J itu d(iw1gka 1wik seba11yak 6 % setiap ta/11111 111a11akala si111pana11 Hassan dijm1gka 1wik seba11yak 8 % setiap ta/11111. Pada I April ta/11111 ke bempakah Hassan dapat 111e111beli mma/J terse/mt secara t1111ai?

Answer I Jawapa11:

[4 marks] [ 4 markah]


11 3472/1

9 Lt is given that k + 8, k and k - 2 are the first three terms of a geometric

progression. Diberi bahawa k + 8, k dc111 k - 2 adalali tiga seb11ta11 µertama bagi s1wt11j{IJ1ia11g

geo111etri.

Find Cari

(a) the va lue of k, 11ilai k,

(b) the common rat io of the progression. nisbah sep1111ya jm1ja11g i11i.

Answer I Jawapan:

(a)

(b)

[3 marks] [3 111arkali]

10 An arithmetic progression bas 18 terms. The first term is J 5. Given that the sum of the last 8 terms is 552, fi nd the common difference of the progression.

S11at11 j{IJ1ia11g aritmetik memµ1myai 18 seb11ta11. Seb11ta11 perlama ialolt J 5. Diberi lwsil tambah 8 seb11tc111 ternkhir adalalt 552, cari beza sep1111yajc111jm1g itu.

Answer I Jmvapa11:

[3 marks] [3 markali]

For Exn111i11er's

U1·e

10

Gj

3472/1 (Lihat halaman sebo tutormansor.wordpress.com

12 3472/1

For 11 The points P(l ,-2), Q, R(5,4) and Sare the vertices of a rhombus. Find the equation of the diagonal QS. (4 marks] £xa111i11er 's

Use

11

[8j

12

cm 0

3472/1

Titik-titik P(l ,-2), Q, R(5 ,4) da11 S adalah bucu-bucu bagi sebuah rombus. Cari persamaa11 pepe11juru QS. [ 4 markah]

Answer I Jmvapa11:

12 After finishing his lesson, Amin looked at the clock on the wall. He thought he could find the equation of the movement of the minute hand if he knew the length of the minute hand from the centre. Selepas selesai pembelqjarc111, Amin melilwt jam di di11di11g. Dia te1jikir dia boleh me11cari persamaa11 pergerakan }arum mini! jika dia tahu pa11ja11g }arum mini! daripada pusat.

Please help Amin to determine the equation of the movement of the minute hand if the length is 5 cm from the centre. (2 marks] To/011g ba11tu Amin me11e11tukc111 persamaa11 pergeraka11 }arum mini! jika panja11g11ya ia/ah 5 cm daripada pusat. (2 111arkah]

Answer I Jawapa11:


13 3472/1

13 Diagram 11 (a) shows the curve y = - 2x3 + 4. Diagram l l(b) shows the straight line graph obtained when y = - 2x3 + 4 is expressed in the linear fo rm Y= 4X + c

14

Rajah 11 (a) 111en1111j11kka11 fengk1111g y = - 2x3 + 4. Rajah 11 (b) 111e111mj11kka11 grqf garis /urns yang diperoleh apabila y = - 2 x3 + 4 di1111gkap dala111 be11t11k linear Y = 4X + c.

y

0

Diagram 11 (a) Raiah 11 (a)

Express X and Y in terms of x and I mor y Ungkapkan X dc111 Y dala111 seb11ta11 x da11 I at au y.

Answer I Jawapa11 :

y

Diagram I l (b) Raiah 11 (b)

x

(3 marks] (3 111arkah]

Given ABCD is a parallelogram, where BC = t-2L and CD= - 3£ + 4L. Find the

unit vector in the direction of AC.

Diberi ABCD ialah segi empat selari, denga11 keadaa11 BC = l - 2j dan

CD = -3£ + 4j . Cari vector 1111it da/0111 arah AC.

Answer I Jawapa11:


For T:xmniner 's

Use

13

cm

14

cm 3472/1 [Lihat lrnlaman sebclah


For 15 Examiner 's

Use

15

cm 0

3472/1

14

Diagram 15 shows the graph of y = 2 cos 11x. Rajah 15 me111111j11kka11 graf y = 2 cos 11x.

y

k

0

- k

(a) State the value of Nyatakan 11ilai

(i) k,

(ii) 11.

Diagram 15 Rajah 15

3472/1

y = 2 cos nx.

(b) Hence, on the axes in the answer space, sketch the graph of y = 12 cos nx - ti

for 0 ~ X ~ 7L

Seter11s11ya, pada paksi yang diberi da/am r11a11g jawapa11, lakarka11 graf

y = 12 cos nx - ll 1111t11k 0 ~ x ~ n.

Answer I Ja111apa11:

(a) (i)

(ii)

(b) y

0



15 3472/1

16 g_ = 2! + 3y

Q. = 4! - Y £ = h ! + ( k + 3h )y

where h and k are constants de11ga11 keadacm h da11 k ialah pemalar

Use the above information to find the value of hand of k if 2 a= 3b - 4c. - - -

Gu11akc111 maklumat ya11g di atas w1tuk 111e11cari 11ilai '1 dan nilai k jika 2a = 3b-4c. - - -

Answer I Jawapa11:


17 The equation of a curve which passes through the point M is y = (x - 3)(x + 2).

The gradient of the normal to the curve at point Mis - ! . 3

Find the coordinates of M. Persa111aa11 s11at11 le11gku11g yang 111ela/11i satu titik M ialah y = (.r - 3)(x + 2).

Kecenman 11or111al kepada lengku11g it11 pada titik M ialah - ! . Cari koordi11af M. 3

Answer I Jmvapcm:

(3 marks] (3 111arkali]

For t:xa111i11er 's

Use

16

cm

17

cm 0

3472/1 [Lihat halaman sebeJah


For 18 E.w1111i11rr's

Use

!• 18

,,; [8j /

0 3472/1

16 3472/l

A circle that is formed from an iron wire has a diameter of I 0 cm. When the wire is heated, its radius increases by 0.04 cm.

Seb11ah b11/ata11 yang di/Jasilka11 daripada suat11 logct111 111e111p1111yai diameter 10 cm. Apab;/a dipa11aska11 jejari b11/ata11 terseb11t bertambah seba11yak 0.04 cm.

Find Cari

(a) the small change in the area of the circle, pernba/Jan kecil pada /11as b11/atc111 terse/mt,

(b) the approximate value of the new area of the circle. a11ggara11 nilai bagi luas b11/ata11 yang barn.

Answer I Jawapa11:

(a)

(b)

[ 4 marks] [ 4 mark ah]


I

I

17 3472/1

19 Diagram 19 shows the shaded region bounded by the curve y = h(x) and the straight linex = 7. Rqjah 19 me11111!i11kka11 rn11ta11 berlorek yang dibatasi oleh le11glw11g y = h(\) da11 garis /11ms x = 7.

y

Diagram 19 Rajah 19

.l' = /i(.r)

It is given that the area of the shaded region is 8 unit2.

Diberi balwwa /11as ra11ta11 berlorek ialah 8 1111it1.

Find Cari

(a)

(b)

11 J h(x) dr, 7 11 J [x + 4/i(x)] dx 7

Answer I Jawapm1:

(a)

(b)


For 1:·.rn111i11er 0.1·

Use

19

[Ej 0

3472/ 1 [Lihat halmnan sebclah


For 20 Hw1111i11er 's

Use

20

EJ 0

3472/1

18 3472/1

Diagram 20 shows a right angled triangle ABC and sector XCY with centre C. Rajah 7.0 111e11111!f11kka11 segi tiga bersuc/111 tepat ABC da11 sector XCY berpusat C.

x

Diagram 20 Rajah 20

ll is given that AB = BC = 12 cm and the ratio CY: CB = 3 : 4. Diberi bahmva AB = BC = 12 cm da1111isba/J CY: CB = 3 : 4

[Use I G1111a n = 3.142]

r ind Cari (a) L. ACB in radians. Give your answer correct to 4 decimal places.

L. A CB da/a1JJ radian. Beri jmvapa11 anda betul kepada 4 te111pat perp11/11/Ja11.

(b) the area, in cm2, of the shaded region.

luas, clala111 cm2, kmvasa11 berlorek.

Answer I Jawapa11:

(a)

(b)

[4 marks] [ 4 111arka/J]


19 3472/1

21 Table 2 1 shows the frequency distribution of the talk time during a call of 40 people. .Jadual 21 111e1111nj11kkm1 tab11rn11 kekernpttn masa bercakap da/am telefo11 bagi 40 orn11g.

Talk tim~ (minutes) Masa Bercakap (minit)

3.00 - 3.40

3.50 - 3.90

4.00 - 4.40

4.50 - 4.90

5.00 - 5.40

5.50 - 5.90

Table 21 Jad11al 21

Find the median of the talk time. Cari 111edia11 bagi masa bercakap.

Answer I Jawapc111:

Number of people Bi/a11ga11 orang

6

4

7

8

12

3

[3 marks] [3 111arka'1]

For E.rn111i11l'r 's

Use

21

[Jij 0

3472/1 ILihat halaman sebelah


For Examiner s

Use

22

cm 0

3472/1

22

20 3472/1

A class of 30 st11dents took a special test. Their scores are shown in Table 22. Seb11alr kelas yang terdiri daripada 30 orang pe/ajar 111e11ga111bi/ 11jia11 klras. Kep11!11sm1 skor 111ereka adalalr seperti da/0111 Jad11al 22.

Score skor

I

2

3

4

5

(a) State the modal score. 1\01afalw11 skor 111od.

Number of students Bilc111ga11 pelqjar

Table 22 Jad11a/ 22

9

6

5

2

8

(b) Calculate the standard deviation. Hit1111g sisilrm1 piawai.

A nswcr I Ja111apa11:

(a)

(b)

(4 marks] [ 4 markalr]


23

21 3472/1

There are 6 different flavours of ice cream : Carrot, Mango, Honey Dew, Van illa, Strawberry and Chocolate. Terdapal 6 perisa ais krim yang berbeza : Lobctl<, Ma11gga, Tembilmi Susu, Va11i/a, Strawberi da11 Coklat.

Find Cari

(a) the number of ways 3 different flavours of ice cream can be chosen, bila11gan cam 3 perisa ais krim yang berbeza bole/, dipilih,

(b) the number of ways at least 5 different navours of ice cream can be chosen. hila11ga11 cara sekurang-kura11g11ya 5 perisa ois krim yang berbeza boleli dipilili .

Answer I Jawapa11:

(a)

(b)

[3 marks] [3 111arka/J]

For li.rn111i11er "s

Us"

23

cm 0

3472/1 ILihat halaman scbelah


fo,· 24 Examiner's

Use

24

[Jij

0 3472/1

22

Two fair dice are rolled together. Dua dad11 adil dila111b1111g bersa111a.

Find the probability that the numbers on them Cari kebara11gka/ia11 balta1va 110111bor pada dad11 itu

(a) have a sum of 11 111e111p11nyai hasil ta111bah 11

(b) have at least one number '6 ' 111e111p1111yai sek11ra11g-k11ra11g11ya satu 110111bor '6 '.

Answer I Jawapa11:

(a)

(b)

3472/1

[2 marks] [2 111arkah]


23 34~1211

25 In a game of guess ing, the probabi lity of guessing correctly is p. The mean and the

standard deviation of success are 15 and l15 respectively. 2

Dala111 per111ai11a11 tekaan, kebara11gkalim1 tekaa11 bet11/ ialah p. Min dan sisiha11 piawai

kejayaan 111asing-111asi11g adalah 15 clan "}__ J5 . 2

Find Cari

(a) the value of p, 11ilai p,

(b) the number of trials required. bilangan c11baa11 yang diperluka11.

Answer I Jmvapa11:

(a)

(b)

END OF QUESTION PAPER KERTASSOALAN TAMAT

[4 marks] [ 4 111arkah]

For Exa111i11er 0s

Use

25

[8j

3472/1 ILihat halaman sebelah


24

INFORMATION FOR C ANDIDATES MA/(LUMAT UNTU/( CALON

1. This question paper consists of 25 questions. Ker/as soalan ini mengand11ngi 25 soa/a11.

2. Answer all questions. Jm vab semua soala11.

3. Write your answers in the spaces provided in this question paper. .Jawapa11 hendaklah dit11/is pada ma11g ya11g disediaka11 dalam kertas soalan ini.

4. Show yom working. It may help you to get marks.

3472/1

T111!j11kkm1 langkah-la11gkah pent ing do/al/I kerja l/lengira 011da. !iii boleh l/le111ba11t11 anda 11nt11k mendapatlwn 111arkali.

5. If you wish to change your answer, cross out the work that you have done. Then write clown the new answer. Sekiranya anda he11dak 111en11kar ja111apa11, batalkm1 jmvapa11 yang telah dib11at. Ke1n11dia11 t11/isjawapa11 yang bam.

6. The diagrams in the questions provided are not drawn to scale unless stated. Rajah yang me11giri11gi soalan tidak di/11kis mengik11t ska/a kec11ali dinyataka11.

7. The marks allocated for each question are shown in brackets. Markah yang dipe/'l/11/11kkan bagi setiap soalan dit111u'11kkan dalmn lwmngan.

8. The Upper Tail Probability Q(z) For The Normal Distribution N(O, 1) Table is provided on page 2. Jad11al Kebarangkalian f-111j11ng Alas Q(z) Bagi Tab11ran Normal N(O, 1) disediakan di

hala111an 2.

9. A list of formulae is provided on pages 3 to 5. Sat11 se11C1mi /'I/JI/II.\' disediakan di halamm1 3 hingga 5.

10. You may use a sci en ti fi e ca l cu la tor. Anda dibe11arka11 me11gg1111alw11 kalk11/ator suint({tk.

11. Hanel in thi s question paper to the invigilator at the end of the examination. Semhkan kerfas soalan ini kepada pengmvas peperiksaan di akhir peperiksaan.

3472/1 ILihat hala man sebela h


I '

ADDITJONAL MATHEMATICS Kertas I Ogas 2 jam

MAJLIS PENGETUA SEKOLAH MALAYSIA CA WANGAN PULAU PINANG

MODUL LArflHAN BERFOKUS SPM 2015

MARJ( SCHEME

ADDITIONAL MATHEMATICS

Paper 1

Two hours


2 347'2/J

ADDITIONAL MATH EMJ\TICS PA Pl~H I 2 01 ~

-Ques tion Solutio11 and Mark Schc 111 c Sub Total

Marks Mark

1 (a) Many lo one 1

I 2 (b) { c, g}

2 3 2 2

Bl: -3!!!_ = 6 g - I (x) = 2111 + mx or 4-m x

3 (a) 3 I

. I 2

(b) 1 ..,

>--·

4 6 3 3

B2: (-6/ - 4(1)(3 + t) = 0 or y = - (x - 3)2 + 6

B I: x2 - 6x + 3 + I = 0 or y = - [x2 - 6x+(-3/ - (-3)2]-3

5 (a) m = 1 and n = 8 2

Bl: m = l

(b) (1, -8) 1

(l, -4) 1 4

6 3 3 3

B2: 4 +x + 2 = 3x

Bl : 16 = 24 or 8 = 23x

3472/1 tutormansor.wordpress.com

~--

7

8

9 (a)

(b)

3472/1

3

2k - 3 - " 2

I o r -- (2k - 3 - h) ci 1 k

2

3 h

2 2

2 log3 y - log3 27 - log3 x B3 : ~--"-~~-=--~~--'--

2log3 y log3 27 log3 x or - --- log3 9 Jog3 9 log3 9

Jog3 y 2 - log 3 27 x

B2 : log3 9

or log 9 y2

- log 9 27 - log 9 x

or log9 / - log9 27x

2024

B3: n = 11

B2: 100 000(1.08)1'- 1 > 120 OOO(l.06y-1

131 : a = 120000, r = 1.06 or a = 100000, r = 1.08 or

OR 2024

133 : n = 11

B2: 120000,127200, 134832, 142922, 151497, 160587, 170222' 180435' 191262, 202737, 214902 and 100000' 108000, 116640, 125971, 136049' 146933., 158687' 171382, 185093' 199900, 215892

Bl : 120000 ,127200' 134832, ... 01'

100000 ' 108000, 116640, .. .

8 -3

Bl: k k - 2 - ----

k+8 k

1 -4

3472/1 -

4 4

4 4

2

1 3


--- - --- - - Jtl7 211 -~ - - -

1 u 4 3 ~

n 2 : ~(2( 1 5 ·I IOd)+ 7d)= 552 2

or ~[2{1 5)+ (18- 1 )ct ) - .!_Q [2{15)+ (1 0 - I }I] = 552 2 2

18 ~[2(1 5)+(1 0 - l )ct ] Bl : T, 1 = 15 + I Od or - [2(1 5)+ (1 8 - l)ct] or 2 2

11 3y = - 2x + 9 2 4 4 or y = - - x + 3 3

B3: y -* l = * - ~ (x - * 3) or *I = *- j(*3) +c

B2: (3, 1) and 2

lllQS = - -3

Bl: (3, 1) 3 or 111PR - -

2

12 x2 + y 2 - 25 = 0 2 2

Bl : ,J (x - 0) 2 + (y ~ 0) 2 = 5

· Note : Accept any coordinates for the centre.

13 Y = L X =-1

x3 )

X J 3 3

B2: Y = L 1 or· X =-

X J X3

Bl: y 4 -=-2+ -x3 x3

14 -

1- (4i - 6j )

3 3

152

B2: ~(4)2 +(- 6)2

~ AC{46J Bl: AC = 4i - 6j or


5 3472/J --- -

15 (a) ( i) k = 2

(ii ) II = 3

(b) )'

3 - ---- -- - 2 4

0 7r

Bl : Y

16 h = 2 and

- 33 k = - or k = - 8.25 3 3

4

B2.: 4 = 12 - 4h or 6 = - 3 - 4k - l 2h

Bl : 2GJ ~ 3( ~i)- 4(k:13h J 17 (2, -4) 3 3

B2: x = 2

Bl: dy = 2x-l dx

or 1111 = 3

18 (a) 2 0.4n 1.257

2 -Tr or or 5

Bl : b'A = lOn x 0.04

(b) 25.4 7r 127

79.80 or - 7( or 2 4 5

Bl : Ao = 25n


6 3472/l -- - ------ --

19 ( ll) -8 J

3 4 (b) 4

8 2: [ 11 2 72

] 2-2 +4(-8)

Bl: x i 2 Q[__ 4(- 8)

20 (a) 0.7855 l

3 4 (b) 40.19

B2 : _!_x 12x 12 - ]_(9)2 * (0.7855) 2 2

Bl : 1 _!_ (9)2 (0. 7854). -x 12 x 12 or

2 2

21 4.638

371 3 3 or -80

. . [ 40 17 J B2 : . 4.45+ 2

8 0.5

Bl : L = 4.45orF=l7orf=8or c = 0.5

22 (a) 1 1

(b) 1.579 3 4

B2: 1310 - (2.8)2

' 30 -

Bl : x =2.8


7 3472/J - - - - - - --~ ~- -

23 (a) 20 I

(b) 7 2 3

Bl: r.c + 6C 5 6 01' 6Cs + I or 6+ 6C 6

24 (a) I I -18

(b) I I l 2 -36

25 (a) l - 3 4

B2: 3 q=-4

Bl: np = 15 or Jl5q =1_.Js 2

(b) 60 I 4


TJNGKATAN CAW ANG AN PU LAU PIN ANG - TUTOR MANSOR · nama tjngkatan maj lis pengetua sekolah...

Documents

Transcript of TJNGKATAN CAW ANG AN PU LAU PIN ANG - TUTOR MANSOR · nama tjngkatan maj lis pengetua sekolah...